Answer:
Explanation:
[tex]496=\frac{480\times 60}{demand}[/tex]
demand per day = 58 pans
Due to availability of two workers we can have parallel we can have deep drawing and trimming operations simultaneously.
Hence the cycle time would be the greater time of the two operations.
cycle time = 450 seconds
[tex]\text{capacity of work cell}=\frac{\text{available working time}}{\text{cycle time}}[/tex]
[tex]\text{capacity of work cell}=\frac{480\times 60}{450}[/tex]
[tex]\text{capacity of work cell}=64 ~pans[/tex] (which is greater than the demand of 58 pans)
Therefore the work cell has sufficient capacity and time (496 sec.>cycle time 450 sec) to meet the demand.
b)
Required daily production is 58 pans
An ideal neon sign transformer provides 9130 V at 51.0 mA with an input voltage of 240 V. Calculate the transformer's input power and current.
Answer:
Input power = 465.63 W
current = 1.94 A
Explanation:
we have the following data to answer this question
V = 9130
i = 0.051
the input power = VI
I = 51.0 mA = 0.051
= 9130 * 0.051
= 465.63 watts
the current = 465.63/240
= 1.94A
therefore the input power is 465.63 wwatts
while the current is 1.94A
the input power is the same thing as the output power.
The system is initially moving with the cable taut, the 15-kg block moving down the rough incline with a speed of 0.080 m/s, and the spring stretched 39 mm. By the method of this article, (a) determine the velocity v of the block after it has traveled 99 mm, and (b) calculate the distance d traveled by the block before it comes to rest.
Solution :
The spring is expanded by 2 times of the block when it moves down an inclined by x times.
Here, [tex]$x_1$[/tex] = 39 mm
[tex]x_2[/tex] = 225 mm
a). From the work energy principal,
Work forces = kinetic energy
[tex]$(mg \sin 50^\circ)\times \frac{99}{1000}-(\mu_k mg \cos 50^\circ) \times \frac{99}{1000} -\frac{1}{2}k(0.225^2 - 0.039^2)=\frac{1}{2}m(V^2_2-0.08^2)$[/tex]
[tex]$(112.6 \times 0.099)-(14.17 \times 0.099)-4.91= 7.5(V^2_2-0.08^2)$[/tex]
[tex]$9.75= 7.5(V^2_2-0.08^2)$[/tex]
[tex]$1.3= V^2_2-0.08^2$[/tex]
[tex]$V_2=1.14\ m/s$[/tex]
b). calculating the distance travelled by the block before it comes to rest.
Substitute the value of [tex]V_2[/tex] in (1),
[tex]$-(\mu_kmg \cos 50^\circ)x + (mg \sin 50^\circ)x-\frac{1}{2}k\left( ( 2x+0.039)^2 - 0.039^2\right)= -\frac{1}{2}m(0.08)^2$[/tex]
[tex]$-14.17x+112.6x - 100(4x^2+0.156x)=-0.048$[/tex]
[tex]$98.43x - 100(4x^2+0.156x)+0.048=0$[/tex]
[tex]$98.43x - 400x^2-15.6x+0.048=0$[/tex]
[tex]$82.83x - 400x^2+0.048=0$[/tex]
[tex]$ 400x^2- 82.83x-0.048=0$[/tex]
x = 0.20 m
Select the correct statement(s) regarding network physical and logical topologies.
a. While logical topologies can be configured in star, ring, bus, and tree configurations, the physical topology must always be in a full-mesh topology
b. logical topologies always incorporate centralized access, whereas physical topologies are always configured as a distributed access network
c. the physical topology addresses how devices are connected, while a logical topology addresses how devices actually communicate to one another
d. all statements are correct
Answer:
The physical topology addresses how devices are connected, while a logical topology addresses how devices actually communicate to one another ( C )
Explanation:
Network physical is simply the process/method of connecting the Network using cables while Logical topology is the general architecture of the communication mechanism in the network for all nodes.
Hence The correct statement is the physical topology addresses how devices are connected, while a logical topology addresses how devices actually communicate to one another
A work-mode-choice model is developed from data acquired in the field in order to determine the probabilities of individual travelers selecting various modes. The mode choices include automobile drive-alone (DL), automobile shared-ride (SR), and bus (B). The utility functions are estimated as follows:
UDL = 2.6 - 0.3 (costDL) - 0.02 (travel timeDL)
USR = 0.7 - 0.3 (costSR) - 0.04 (travel timeSR)
UB = -0.3 (costB) - 0.01 (travel timeB)
where cost is in dollars and time is in minutes. The cost of driving an automobile is exist5.50 with a travel time of 21 minutes, while the bus fare is exist1.25 with a travel time of 27 minutes. How many people will use the shared-ride mode from a community of 4500 workers, assuming the shared-ride option always consists of three individuals sharing costs equally?
a. 314
b. 828
c. 866
d. 2805
Answer:
b. 828
Explanation:
UDL = 2.6 - 0.3 [5.5] - 0.02 [ 21 ] = 0.53
USR = 0.7 - 0.3 [5.5 / 2 ] - 0.04 [ 21 ] = -0.69
UB = -0.3 [ 1.25 ] - 0.01 = -0.645
Psr = [tex]\frac{e^{-0.53} }{e^{-0.53} + e^{-0.69} + e^{-0.645} }[/tex]
Solving the equation we get 0.184.
Number of people who will take shared ride is:
0.184 * 4500 = 828 approximately.
A 20-mm-diameter steel bar is to be used as a torsion spring. If the torsional stress in the bar is not to exceed 110 MPa when one end is twisted through an angle of 15°, what must be the length of the bar?
Answer:
1.887 m
Explanation:
(15 *pi)/180
= 0.2618 rad
Polar moment
= Pi*d⁴/32
= (22/7*20⁴)/32
= 15707.96
Torque on shaft
= ((22/7)*20³*110)/16
= 172857.14
= 172.8nm
Shear modulus
G = 79.3
L = Gjθ/T
= 79.3x10⁹x(1.571*10^-8)x0.2618/172.8
= 1.887 m
The length of the bar is therefore 1.887 meters
Water is pumped steadily through a 0.10-m diameter pipe from one closed pressurized tank to another tank. The pump adds 4.0 kW of energy to the water and the head loss of the flow is 10 m. Determine the velocity of the water leaving the pump and discharging into tank B.
Complete Question
Complete Question is attached below.
Answer:
[tex]V'=5m/s[/tex]
Explanation:
From the question we are told that:
Diameter [tex]d=0.10m[/tex]
Power [tex]P=4.0kW[/tex]
Head loss [tex]\mu=10m[/tex]
[tex]\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu[/tex]
[tex]\frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10[/tex]
[tex]H_m=(\frac{200*10^3}{1000*9.8}-10)[/tex]
[tex]H_m=10.39m[/tex]
Generally the equation for Power is mathematically given by
[tex]P=\rho gQH_m[/tex]
Therefore
[tex]Q=\frac{P}{\rho g H_m}[/tex]
[tex]Q=\frac{4*10^4}{1000*9.81*10.9}[/tex]
[tex]Q=0.03935m^3/sec[/tex]
Since
[tex]Q=AV'[/tex]
Where
[tex]A=\pi r^2\\A=3.142 (0.05)^2[/tex]
[tex]A=7.85*10^{-3}[/tex]
Therefore
[tex]V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}[/tex]
[tex]V'=5m/s[/tex]
what is the best glide speed for your training airplane
1.5 nautical miles per 1,000 feet
Problema:
Una nevera de vinos, con un peso bruto de 50 kg., que tiene las siguientes dimensiones: .60 m Largo x .49 m ancho x .50 m altura. Para ser transportadas en un contenedor de 40 pies D.V. responder las siguientes preguntas:
• 1.Cuántas neveras de vinos de acuerdo al volumen caben en un contenedor de 40 pies?
• De acuerdo dimensiones internas (largo, ancho y alto), ¿Cuántas caben en un contenedor de 40 pies?
• De acuerdo al peso que soporta el contenedor. ¿Cuántas neveras de vinos es posible transportar?
Answer:
I can't understand this language .
Identify the first step in preparing a spectrophotometer for use.
A. Make sure all samples and the blank are ready for measurement.
B. Prepare a calibration curve.
C. Measure the absorbance of the blank.
D. Turn on the light source and the spectrophotometer.
Answer:
D. Turn on the light source and the spectrophotometer.
Explanation:
A spectrophotometer is a machine used to measure the presence of any light-absorbing particle in a solution as well as its concentration. To prepare a spectrophotometer for use, the first step is to turn on the spectrophotometer and allow it to warm up for at least 15 minutes. After this is done, the next step will be to ensure that the samples and blank are ready. Next, an appropriate wavelength is set for the solute being determined. Finally, the absorbance is measured of both the blank and samples.
3-71A 20mm diameter steel bar is to be used as a torsion spring. If the torsional stress in the bar is not to exceed 110 MPa when one end is twisted through an angle of 15 degrees, what must be the length of the bar
Answer:
The right answer is "1.903 m".
Explanation:
Given that,
[tex]\tau =110 \ MPa[/tex]
[tex]G=80 \ GPa[/tex]
[tex]\Theta=15\times \frac{\pi}{180}[/tex]
[tex]=\frac{\pi}{12}[/tex]
[tex]d=20 \ mm[/tex]
As we know,
⇒ [tex]\frac{\tau}{r}=\frac{G \Theta}{L}[/tex]
Or,
⇒ [tex]L=\frac{G \theta r}{\tau}[/tex]
[tex]=\frac{80\times 10^3}{110}\times \frac{\pi}{12}\times 10[/tex]
[tex]=1903.9 \ mm[/tex]
or,
[tex]=1.903 \ m[/tex]
Determine the pressure difference in N/m2,between two points 800m apart in horizontal pipe-line,150 mm diameter, discharging water at the rate of 12.5litres per second. Take the frictional coefficient ,f, as being 0.008
Answer: [tex]10.631\times 10^3\ N/m^2[/tex]
Explanation:
Given
Discharge is [tex]Q=12.5\ L[/tex]
Diameter of pipe [tex]d=150\ mm[/tex]
Distance between two ends of pipe [tex]L=800\ m[/tex]
friction factor [tex]f=0.008[/tex]
Average velocity is given by
[tex]\Rightarrow v_{avg}=\dfrac{12.5\times 10^{-3}}{\frac{\pi }{4}(0.15)^2}\\\\\Rightarrow v_{avg}=\dfrac{15.9134\times 10^{-3}}{2.25\times 10^{-2}}\\\\\Rightarrow v_{avg}=7.07\times 10^{-1}\\\Rightarrow v_{avg}=0.707\ m/s[/tex]
Pressure difference is given by
[tex]\Rightarrow \Delta P=f\ \dfrac{L}{d}\dfrac{\rho v_{avg}^2}{2}\\\\\Rightarrow \Delta P=0.008\times \dfrac{800}{0.15}\times \dfrac{997\times (0.707)^2}{2}\\\\\Rightarrow \Delta P=10,631.45\ N/m^2\\\Rightarrow \Delta P=10.631\ kPa[/tex]
A 1m3 tank containing air at 25℃ and 500kPa is connected through a valve to
another tank containing 5kg of air at 35℃ and 200kPa. Now the valve is opened,
and the entire system is allowed to reach thermal equilibrium, which is at 20℃
(Take: Ru = 8.314 kJ / kg.K).
Answer:
The right answer is "2.2099 m³".
Explanation:
Given:
Mass,
m = 5 kg
Temperature,
T = 35℃
or,
= 35 + 273
Pressure,
P = 200 kPa
Gas constant,
R = 0.2870 kj/kgK
By using the ideal gas equation,
The volume will be:
⇒ [tex]PV=mRT[/tex]
or,
⇒ [tex]V=\frac{mRT}{P}[/tex]
By substituting the values, we get
[tex]=\frac{5(0.2870)(35+273)}{200}[/tex]
[tex]=\frac{441.98}{200}[/tex]
[tex]=2.2099 \ m^3[/tex]
In an international film festival, a penal of 11 judges is formed to judge the best film. At
last two films FA and FB were considered to be the best where the opinion of judges got
divided. Six judges where in favor of FA whereas five in favor of FB. A random sample
of five judges was drawn from the panel. Find the probability that out of five judges,
three are in favor of film FA.Enunciate demerits of classical probability.
Answer:
International Film Festival
Judging the best best film:
a. The probability that out of five judges (random sample), three are in favor of film FA is:
= 33%.
b. The demerits of classical probability are:
1. Classical probability can only be used with events that have definite numbers of possible outcomes.
2. Classical probability can only handle events where each outcome is equally likely.
3. Classical probability is based on the assumption of linear relationship (which is not always true in real life) between the latent variable and observed scores.
Explanation:
a) Number of judges = 11
Number of judges in favor of FA film = 6
Number of judges in favor of FB film = 5
Probability of judges in favor of FA film = 6/11
Probability of judges in favor of FB film = 5/11
Random sample of judges = 5
Probability that out of five judges, three are in favor of film FA = 3/5 * 6/11
= 18/55
= 33%
b) Classical probability is the simple probability showing that each event has equal chance of happening. It can be contrasted with empirical probability that is obtained from experiments.
(50 POINTS) How many people use pipes in the world? How do you know this?
Answer:
7.9 billion people
Explanation:
) Please label the following statements as either True (T) or False (F). (a) In general, the greater the % of cold work, the smaller the recrystallization grain size. (b) The higher the annealing temperature, the smaller the recrystallization grain size. (c) The greater the % of cold work, the lower the recrystallization temperature.
Answer:
A. This option is true
B. This option is false
C. This option is true
Explanation:
A. Generally speaking, the greater percentage of cold, the recrystallization grain size would turn out to be smaller. Therefore this true.
B. A higher annealing temperature does not result in smaller recrystallization grain size. Therefore this is false.
C. As the percentage of cold work is greater, the recrystallization temperature would tend to be lower. Therefore this is true.
The following is a correlation for the average Nusselt number for natural convection over spherical surface. As can be seen in the above, the Nusselt number approaches 2 as Rayleigh number approaches zero. Prove that this situation corresponds to conduction heat transfer and in conduction heat transfer over sphere, the Nusselt number becomes 2. Hint: First step: Write an expression for heat transfer between two spherical shells that share the same center. Second step: Assume the outer spherical shell is infinitely large.
Answer:
Explanation:
[tex]r_2=[/tex]∞
[tex]q=4\pi kT_1(T_2-T_1)\\[/tex]
[tex]q=2\pi kD.[/tex]ΔT--------(1)
[tex]q=hA[/tex] ΔT[tex]=4\pi r_1^2(T_2_s-T_1_s)\\[/tex]
[tex]N_u=\frac{hD}{k} = 2+\frac{0.589 R_a^\frac{1}{4} }{[1+(\frac{0.046}{p_r}\frac{9}{16} )^\frac{4}{9} } ------(3)[/tex]
By equation (1) and (2)
[tex]2\pi kD.[/tex]ΔT=h.4[tex]\pi r_1^2[/tex]ΔT
[tex]2kD=hD^2\\\frac{hD}{k} =2\\N_u=\frac{hD}{k}=2\\[/tex]-------(4)
From equation (3) and (4)
So for sphere [tex]R_a[/tex]→0
The temperature gradient in a spherical (or cylindrical) wall at steady state will always decrease (in magnitude) with increasing distance from the center (line), i.e. radial distance.
A. True
B. False
Answer:
True
Explanation:
Yes it is true that the Temperature gradient would also decrease with magnitude just as the distances rise from the centre line.
We have this cylinder equation as
[T1-T2 / ln(r1-r2)]2πKL
The radial distance is r2-r1
The gradient of temperature is T1-T2
From the equation,
The temperature gradient has a direct and proportional relationship to radial distance
T1-T2 ∝ ln(r2-r1)
1/T1-T2 = k(r2-r1)
This inverse relationship above confirms that the statement is true
Set the leak rate to zero and choose a non-zero value for the proportional feedback gain.Restart the simulation and turn on the outflow valve.What happens to the liquid level in the tank?Repeat this process with higher and lower values for the proportional feedback gain.What happens when the proportional feedback gain is increased?What happens when it is decreased?Find the proportional gain that will reach steady state the quickest without oscillationin the state of the valve and restart the simulation.What is the system time constant, as determined from the tank level versus time plot.
Answer:
Explanation:
The proportional gain K is usually a fixed property of the controller . If proportional gain is increased , The sensitivity of the controller to error is increased but the stability is impaired. The system approaches the behaviour of on off controlled system and it response become oscillatory
A horizontal water jet impinges against a vertical flat plate at 30 ft/s and splashes off the sides in the verti- cal plane. If a horizontal force of 500 lbf is required to hold the plate against the water stream, determine the volume flow rate of the water.
Answer:
8.6 ft³/s
Explanation:
The force due to the water jet F = mv where m = mass flow rate = ρQ where ρ = density of water = 62.4 lbm/ft³ and Q = volume flow rate. v = velocity of water jet = 30 ft/s
So, F = mv
F = ρQv
making Q subject of the formula, we have
Q = F/ρv
Since F = force due to water jet = force needed to hold the plate against the water stream = 500 lbf = 500 × 1 lbf = 500 × 32.2 lbmft/s² = 16100 lbmft/s²
Since
Q = F/ρv
Substituting the values of the variables into the equation for Q, we have
Q = F/ρv
Q = 16100 lbmft/s²/(62.4 lbm/ft³ × 30 ft/s)
Q = 16100 lbmft/s²/1872 lbm/ft²s
Q = 8.6 ft³/s
So, the volume flow rate is 8.6 ft³/s.
In low speed subsonic wind tunnels, the value of test section velocity can be controlled by adjusting the pressure difference between the inlet and test-section for a fixed ratio of inlet-to-test section cross-sectional area.
a. True
b. false
Answer:
Hence the given statement is false.
Explanation:
For low-speed subsonic wind tunnels, the air density remains nearly constant decreasing the cross-section area cause the flow to extend velocity, and reduce pressure. Similarly increasing the world cause to decrease and therefore the pressure to extend.
The speed within the test section is decided by the planning of the tunnel.
Thus by adjusting the pressure difference won't change the worth of test section velocity.
Answer:
The given statement is false .
A flow inside a centrifuge can be approximated by a combination of a central cylinder and a radial line source flow, giving the following potential function:
Ø= a2/r -cosØ + aßlnr = r
Where a is the radius of the central base of the centrifuge and ß is a constant.
a) Provide expressions for the velocities Vr and vo .
b) Find the expression for the stream function.
Answer:
a) Vr = - a^2/r cosθ + aß / r
Vθ = 1/r [ -a^2/r * sinθ ]
b) attached below
Explanation:
potential function
Ø= a^2 /r cosØ + aßlnr ----- ( 1 )
a = radius , ß = constant
a) Expressions for Vr and Vθ
Vr = dØ / dr ----- ( 2 )
hence expression : Vr = - a^2/r cosθ + aß / r
Vθ = 1/r dØ / dθ ------ ( 3 )
back to equation 1
dØ / dr = - a^2/r sinθ + 0 --- ( 4 )
Resolving equations 3 and 4
Vθ = 1/r [ -a^2/r * sinθ ]
b) expression for stream function
attached below
The heat transfer surface area of a fin is equal to the sum of all surfaces of the fin exposed to the surrounding medium, including the surface area of the fin tip. Under what conditions can we neglect heat transfer from the fin tip?
Answer:
The explanation according to the given query is summarized in the explanation segment below.
Explanation:
If somehow the fin has become too lengthy, this same fin tip temperature approaches the temperature gradient and maybe we'll ignore heat transmission out from end tips.Additionally, effective heat transmission as well from the tip could be ignored unless the end tip surface is relatively tiny throughout comparison to its overall surface.The following measurements are taken on particular junction diodes for which V is the terminal voltage and I is the diode current. For each diode, estimate values of Is and the terminal voltage at 10% of the measured current.
(a) V = 0.700 V at I = 1.00 A.
(b) V = 0.650 V at I = 1.00 mA.
(c) V = 0.650 V at I = 10 mu A.
(d) V = 0.700V at I = 100 mA.
The values of Is and V are as: (a) [tex]Is = 2.34 \times 10^{-11} A[/tex] and V = 0.581 V. (b) [tex]Is = 4.56 \times 10^{-15} A[/tex] and V = 0.516 V. (c) [tex]Is = 1.18 \times 10^{-16} A\\[/tex] and V = 0.459 V. (d) [tex]Is = 2.34 \times 10^{-11} A[/tex] and V = 0.581 V.
The relation between the current and voltage of a diode is given by the Shockley diode equation. It is an exponential function and can be given by the following equation:
[tex]I = Is \times (e^{V/Vt} - 1)[/tex]
Where
I = currentV = voltageVt = thermal voltageIs = reverse saturation current.(a)
Given that:
V = 0.700 V
And I = 1.00 A.
Substituting these values in the equation above to get,
[tex]1.00 A = Is \times (e^{0.700 V / 0.025 V} - 1)\\Is = 2.34 \times 10^{-11} A[/tex]
The terminal voltage at 10% of the measured current can be found by substituting I = 0.1 A in the above equation and solving for V as:
V = 0.581 V.
(b)
Given that:
V = 0.650 V
And, I = 1.00 mA.
Substituting these values in the equation above to get,
[tex]1.00 mA = Is \times (e^{0.650 V / 0.025 V} - 1)\\ Is = 4.56 \times 10^{-15} A[/tex]
The terminal voltage at 10% of the measured current can be found by substituting I = 0.1 mA in the above equation and solving for V as:
V = 0.516 V.
(c)
Given that:
V = 0.650 V
And, I = 10 μA.
Substituting these values in the equation above to get,
[tex]10 \mu A = Is \times (e^{0.650 V / 0.025 V} - 1)\\Is = 1.18 \times 10^{-16} A[/tex]
The terminal voltage at 10% of the measured current can be found by substituting I = 1 μA in the above equation and solving for V as:
V = 0.459 V.
(d)
Given that:
V = 0.700 V
And, I = 100 mA.
Substituting these values in the equation above to get,
[tex]100 \ mA = Is \times (e^{0.700 V / 0.025 V} - 1)\\Is = 2.34 \times 10^{-11} A[/tex]
The terminal voltage at 10% of the measured current can be found by substituting I = 10 mA in the above equation and solving for V as:
V = 0.581 V.
So, the values of Is and V are as: (a) [tex]Is = 2.34 \times 10^{-11} A[/tex] and V = 0.581 V. (b) [tex]Is = 4.56 \times 10^{-15} A[/tex] and V = 0.516 V. (c) [tex]Is = 1.18 \times 10^{-16} A\\[/tex] and V = 0.459 V. (d) [tex]Is = 2.34 \times 10^{-11} A[/tex] and V = 0.581 V.
Learn more about Terminal voltage here:
https://brainly.com/question/34372613
#SPJ3
If a heat engine has an efficiency of 30% and its power output is 600 W, what is the rate of heat input from the combustion phase
Answer:
The heat input from the combustion phase is 2000 watts.
Explanation:
The energy efficiency of the heat engine ([tex]\eta[/tex]), no unit, is defined by this formula:
[tex]\eta = \frac{\dot W}{\dot Q}[/tex] (1)
Where:
[tex]\dot Q[/tex] - Heat input, in watts.
[tex]\dot W[/tex] - Power output, in watts.
If we know that [tex]\dot W = 600\,W[/tex] and [tex]\eta = 0.3[/tex], then the heat input from the combustion phase is:
[tex]\eta = \frac{\dot W}{\dot Q}[/tex]
[tex]\dot Q = \frac{\dot W}{\eta}[/tex]
[tex]\dot Q = \frac{600\,W}{0.3}[/tex]
[tex]\dot Q = 2000\,W[/tex]
The heat input from the combustion phase is 2000 watts.
Steam enters an adiabatic turbine at 6 MPa, 600°C, and 80 m/s and leaves at 50 kPa, 100°C, and 140 m/s. If the power output of the turbine is 5 MW, determine (a) the reversible power output and (b) the second-law efficiency of the turbine. Assume the surroundings to be at 25°C.
Answer:
(a) the reversible power output of turbine is 5810 kw
(b) The second-law efficiency of he turbine = 86.05%
Explanation:
In state 1: the steam has a pressure of 6 MPa and 600°C. Obtain the enthalpy and entropy at this state.
h1 = 3658 kJ/kg s1=7.167 kJ/kgK
In state 2: the steam has a pressure of 50 kPa and 100°C. Obtain the enthalpy and entropy at this state
h2 = 2682kl/kg S2= 7.694 kJ/kg
Assuming that the energy balance equation given
Wout=m [h1-h2+(v1²-v2²) /2]
Let
W =5 MW
V1= 80 m/s V2= 140 m/s
h1 = 3658kJ/kg h2 = 2682 kJ/kg
∴5 MW x1000 kW/ 1 MW =m [(3658-2682)+ ((80m/s)²-(140m/s)²)/2](1N /1kg m/ s²) *(1KJ/1000 Nm)
m = 5.158kg/s
Consider the energy balance equation given
Wrev,out =Wout-mT0(s1-s2)
Substitute Wout =5 MW m = 5.158kg/s 7
s1= 7.167 kJ/kg-K s2= 7.694kJ/kg-K and 25°C .
Wrev,out=(5 MW x 1000 kW /1 MW) -5.158x(273+25) Kx(7.167-7.694)
= 5810 kW
(a) Therefore, the reversible power output of turbine is 5810 kw.
The given values of quantities were substituted and the reversible power output are calculated.
(b) Calculating the second law efficiency of the turbine:
η=Wout/W rev,out
Let Wout = 5 MW and Wrev,out = 5810 kW
η=(5 MW x 1000 kW)/(1 MW *5810)
η= 86.05%
Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not move relative to B. All surfaces are smooth.
Answer:
The answer is "15 N".
Explanation:
Please find the complete question in the attached file.
In frame B:
For just slipping:
[tex]\to \frac{P}{2} \cos \theta =mg \sin \theta\\\\\to P=2 mg \tan \theta \\\\[/tex]
[tex]=2 \times 1 \times g \times \tan 37^{\circ}\\\\ =2 \times 10 \times \frac{3}{4}\\\\ =15 \ N[/tex]
Your shifts productivity is Slow because one person is not pulling his share. The rest of the team is Getting upset.
Answer:
you are right but then you ddnt ask a question
A levee will be constructed to provide some flood protection for a residential area. The residences are willing to accept a one-in-five chance that the levee will be overtopped in the next 15 years. Assuming that the annual peak streamflow follows a lognormal distribution with a log10(Q[ft3/s]) mean and standard deviation of 1.835 and 0.65 respectively, what is the design flow in ft3/s?
Answer:
1709.07 ft^3/s
Explanation:
Annual peak streamflow = Log10(Q [ft^3/s] )
mean = 1.835
standard deviation = 0.65
Probability of levee been overtopped in the next 15 years = 1/5
Determine the design flow ins ft^3/s
P₁₅ = 1 - ( q )^15 = 1 - ( 1 - 1/T )^15 = 0.2
∴ T = 67.72 years
Q₁₅ = 1 - 0.2 = 0.8
Applying Lognormal distribution : Zt = mean + ( K₂ * std ) --- ( 1 )
K₂ = 2.054 + ( 67.72 - 50 ) / ( 100 - 50 ) * ( 2.326 - 2.054 )
= 2.1504
back to equation 1
Zt = 1.835 + ( 2.1504 * 0.65 ) = 3.23276
hence:
Log₁₀ ( Qt(ft^3/s) ) = Zt = 3.23276
hence ; Qt = 10^3.23276
= 1709.07 ft^3/s
6. When the engine stalls or the power unit fails, on a car with power
brakes, the service brake pedal will
A. Take about the same amount of pressure
B. Take more pressure to stop
C. Take less pressure to stop
D. Become locked in place and no longer help stop the car
bending stress distribution is a.rectangle b.parabolic c.curve d.i section
